Question

Here we will calculate the work per unit charge on an electron moving between two potentials. A 9.0 V battery is connected across two large parallel plates that are separated by 4.5 mm of air, creating a potential difference of 9.0 V between the plates. (a) What is the electric field in the region between the plates? (b) An electron is released from rest at the negative plate. If the only force on the electron is the electric force exerted by the electric field of the plates, what is the speed of the electron as it reaches the positive plate? The mass of an electron is me=9.11×10−31kg.

Answer 1

Answer:

2000 V/m

1778021.69323 m/s

Explanation:

V = Voltage = 9 V = $\Delta V$

q = Charge of proton = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

v = Velocity of electron

Electric field is given by

$E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{9}{4.5\times 10^{-3}}\\\Rightarrow E=2000\ V/m$

The electric field is 2000 V/m

Here, the energy of the system is conserved

$\dfrac{1}{2}mv^2=q\Delta V\\\Rightarrow v=\sqrt{\dfrac{2q\Delta V}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 9}{9.11\times 10^{-31}}}\\\Rightarrow v=1778021.69323\ m/s$

The velocity of the electron is 1778021.69323 m/s